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# A ball is dropped from a height of 10 m. If the energy of the ball reduces by 40 % after striking the ground, how high can the ball bounce back? (g = $10m{{s}^{-2}}$)(A) 6 m(B) 10 m(C) 3 m(D) 1 m

Last updated date: 13th Jun 2024
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Hint: We know that in physics, the kinetic energy of an object is the energy that it possesses due to its motion. It is defined as the work needed to accelerate a body of a given mass from rest to its stated velocity. Having gained this energy during its acceleration, the body maintains this kinetic energy unless its speed changes. On the other hand, potential energy is energy that is stored or conserved in an object or substance. This stored energy is based on the position, arrangement or state of the object or substance. We can think of it as energy that has the 'potential' to do work. Based on this concept we have to solve this question.

We know that the total energy of the ball is given as$=\mathrm{m} \mathrm{g} \mathrm{h}$
After we put the value of the height we get as $=\operatorname{mg} 10$
$=100 \mathrm{m} \mathrm{kgm}^{2} \mathrm{s}^{-2}$
It is known that energy with which it moves up the ground after striking = $60 \%$ of the total energy $\mathrm{E}=\dfrac{60}{100} \times 100 \mathrm{m} \mathrm{kgm}^{2} \mathrm{s}^{-2}$
Height to which the ball will bounce back $\mathrm{h}=\dfrac{\mathrm{E}}{\mathrm{m} \times \mathrm{g}}=6$ meter